Class 12: Determinants – Exercise 6.5

$\displaystyle \text{Solve each of the following systems of homogeneous linear equations:}$
$\displaystyle \textbf{Question 1: }x+y-2z=0,\;2x+y-3z=0,\;5x+4y-9z=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the given system of equations, we have}$
$\displaystyle D=\begin{vmatrix}1&1&-2\\2&1&-3\\5&4&-9\end{vmatrix}$
$\displaystyle =1(-9+12)-1(-18+15)-2(8-5)=0$
$\displaystyle \text{Hence, the system has infinitely many solutions.}$
$\displaystyle \text{Putting }z=k\text{ in the first two equations, we get}$
$\displaystyle x+y=2k$
$\displaystyle 2x+y=3k$
$\displaystyle \Delta=\begin{vmatrix}1&1\\2&1\end{vmatrix}=1-2=-1\neq0$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{\begin{vmatrix}2k&1\\3k&1\end{vmatrix}}{\Delta}=\frac{2k-3k}{-1}=k$
$\displaystyle y=\frac{\begin{vmatrix}1&2k\\2&3k\end{vmatrix}}{\Delta}=\frac{3k-4k}{-1}=k$
$\displaystyle \text{Thus, }x=k,\;y=k,\;z=k.$
$\displaystyle \text{Clearly, these values satisfy the third equation.}$
$\displaystyle \therefore x=y=z=k,\quad k\in\mathbb{R}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }2x+3y+4z=0,\;x+y+z=0,\;2x-y+3z=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the given system of equations, we have}$
$\displaystyle D=\begin{vmatrix}2&3&4\\1&1&1\\2&-1&3\end{vmatrix}$
$\displaystyle =2(3+1)-3(3-2)+4(-1-2)$
$\displaystyle =8-3-12=-7\neq0$
$\displaystyle \therefore\text{ the given system has only the trivial solution.}$
$\displaystyle \therefore x=0,\quad y=0,\quad z=0$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }3x+y+z=0,\;x-4y+3z=0,\;2x+5y-2z=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the given system of equations, we have}$
$\displaystyle D=\begin{vmatrix}3&1&1\\1&-4&3\\2&5&-2\end{vmatrix}$
$\displaystyle =3(8-15)-1(-2-6)+1(5+8)$
$\displaystyle =-21+8+13=0$
$\displaystyle \text{Hence, the system has infinitely many solutions.}$
$\displaystyle \text{Putting }z=k\text{ in the first two equations, we get}$
$\displaystyle 3x+y=-k$
$\displaystyle x-4y=-3k$
$\displaystyle \Delta=\begin{vmatrix}3&1\\1&-4\end{vmatrix}=-12-1=-13\neq0$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{\begin{vmatrix}-k&1\\-3k&-4\end{vmatrix}}{\Delta}=\frac{4k+3k}{-13}=-\frac{7k}{13}$
$\displaystyle y=\frac{\begin{vmatrix}3&-k\\1&-3k\end{vmatrix}}{\Delta}=\frac{-9k+k}{-13}=\frac{8k}{13}$
$\displaystyle \therefore x=-\frac{7k}{13},\quad y=\frac{8k}{13},\quad z=k$
$\displaystyle \text{Clearly, these values satisfy the third equation.}$
$\displaystyle \text{Putting }k=13\lambda,\text{ where }\lambda\in\mathbb{R},\text{ we get}$
$\displaystyle \therefore x=-7\lambda,\quad y=8\lambda,\quad z=13\lambda$
$\displaystyle \text{Thus, the system has infinitely many solutions for every }\lambda\in\mathbb{R}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }$
$\displaystyle \text{Find the real values of }\lambda\text{ for which the following system of linear equations has}$
$\displaystyle \text{non-trivial solutions. Also, find the non-trivial solutions.}$
$\displaystyle 2\lambda x-2y+3z=0,\quad x+\lambda y+2z=0,\quad 2x+\lambda z=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations can be written as}$
$\displaystyle 2\lambda x-2y+3z=0$
$\displaystyle x+\lambda y+2z=0$
$\displaystyle 2x+0y+\lambda z=0$
$\displaystyle \text{The given homogeneous system will have non-trivial solutions if }D=0.$
$\displaystyle \therefore\begin{vmatrix}2\lambda&-2&3\\1&\lambda&2\\2&0&\lambda\end{vmatrix}=0$
$\displaystyle \Rightarrow 2\lambda(\lambda^2)+2(\lambda-4)+3(-2\lambda)=0$
$\displaystyle \Rightarrow 2\lambda^3-4\lambda-8=0$
$\displaystyle \Rightarrow 2(\lambda-2)(\lambda^2+2\lambda+2)=0$
$\displaystyle \text{Since }\lambda^2+2\lambda+2=(\lambda+1)^2+1>0\text{ for every real }\lambda,$
$\displaystyle \therefore \lambda=2$
$\displaystyle \text{Now, we find the solutions for }\lambda=2.$
$\displaystyle \text{Putting }z=k\text{ in the first two equations, we get}$
$\displaystyle 4x-2y=-3k$
$\displaystyle x+2y=-2k$
$\displaystyle \Delta=\begin{vmatrix}4&-2\\1&2\end{vmatrix}=8+2=10\neq0$
$\displaystyle \text{By Cramer's rule, we obtain}$
$\displaystyle x=\frac{\begin{vmatrix}-3k&-2\\-2k&2\end{vmatrix}}{\Delta}=\frac{-6k-4k}{10}=-k$
$\displaystyle y=\frac{\begin{vmatrix}4&-3k\\1&-2k\end{vmatrix}}{\Delta}=\frac{-8k+3k}{10}=-\frac{k}{2}$
$\displaystyle \text{Substituting these values in the third equation, we get}$
$\displaystyle 2x+2z=2(-k)+2k=0$
$\displaystyle \text{Hence, these values satisfy the third equation.}$
$\displaystyle \therefore \lambda=2,\quad x=-k,\quad y=-\frac{k}{2},\quad z=k,\quad k\in\mathbb{R},\;k\neq0$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }$
$\displaystyle \text{If }a,\;b,\;c\text{ are non-zero real numbers and the system of equations}$
$\displaystyle (a-1)x=y+z,\quad (b-1)y=z+x,\quad (c-1)z=x+y$
$\displaystyle \text{has a non-trivial solution, prove that }ab+bc+ca=abc.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given system of equations can be written as}$
$\displaystyle (a-1)x-y-z=0$
$\displaystyle -x+(b-1)y-z=0$
$\displaystyle -x-y+(c-1)z=0$
$\displaystyle \text{Since the system has a non-trivial solution, we must have }D=0.$
$\displaystyle \therefore\begin{vmatrix}a-1&-1&-1\\-1&b-1&-1\\-1&-1&c-1\end{vmatrix}=0$
$\displaystyle \Rightarrow (a-1)\big[(b-1)(c-1)-1\big]-\big[c-1+1\big]-\big[1+b-1\big]=0$
$\displaystyle \Rightarrow (a-1)(bc-b-c)-c-b=0$
$\displaystyle \Rightarrow abc-ab-ac-bc+b+c-b-c=0$
$\displaystyle \Rightarrow abc-ab-ac-bc=0$
$\displaystyle \therefore ab+bc+ca=abc$
$\displaystyle \text{Hence proved.}$
$\displaystyle \\$